Physics 504, Physics 504, Spring 2010 The Optical Theorem Spring 2010 Electricity Electricity and The optical theorem relates the total cross section to the and Magnetism forward scattering amplitude. Magnetism Shapiro In Quantum Mechanics, this is “conservation” of Shapiro probability. Here — conservation of energy. Optical Optical Theorem and We do this differently from Jackson. Theorem and Index of Index of Refraction Consider scatterer of finite size in an incident plane wave: Refraction The Optical The Optical Last time we mentioned scattering removes power from Theorem i~ki ~x iωt Theorem E~i = E0 ~i e · − beam. Today we treat this more generally, to find Index of Index of Refraction 1 1 i~k ~x iωt Refraction B~ = ~k E~ = ~k ~ E e i· − I the optical theorem: i ω i × i ω i × i 0 I the relationship of the index of refraction and the Assume scattering is linear, time-invariant physics, so forward scattering amplitude. everything e iωt, make implicit. ∝ − scattering amplitude f~(~k,~ki), so eikr E~ (~x)= f~(~k,~k )E s r i 0 1 eikr B~ (~x)= ~k E~ = E ~k f~(~k,~k ). s ω × s ωr 0 × i

~ ~ Linearity assures frequency unchanged and k = k = ki , Physics 504, Physics 504, | | | | Spring 2010 Spring 2010 elastic scattering. Electricity Electricity and 1 ~ ~ ~ ~ and The total fields are E~ = E~i + E~s and B~ = B~ i + B~ s. Magnetism ∆P = ρdρdφRe Ei Bs∗ + Es Bi∗ Magnetism 2µ0 Z h × × iz Consider the power flowing past planes ~k , one way in Shapiro 2 Shapiro ⊥ i 1 E0 front of the scatterer, one way in back. = | | ρdρdφ Optical 2ωµ0 r Z Optical The total power removed from the incident beam = Theorem and Theorem and ~ Index of ~ ~ ~ ~ ikr+iki ~x Index of incident power flux times σtot, which includes absorption Refraction Re ~i k f ∗(k, ki) e− · Refraction  ×  ×  and scattering cross sections. The Optical The Optical Theorem ~ Theorem ~ ikr iki ~x ~ ~ ~ ~ Take ki zˆ. Total power across z downstream is Index of +e − · f(k, ki) ki ~i ∗ Index of k Refraction ×  × z Refraction 1 ~ ~ ~ ~ P = ρdρdφRe Ei + Es Bi∗ + Bs∗ . ~ ~ 2µ0 Z h  ×  iz For large z, ρ √z so the angle goes to zero, k = ki, ~ ∼ 2 2 2 ~ ~ kr ki ~x = k(r z)=k( z + ρ z) kρ /2z. The Ei Bi∗ part of this is what would have been the − · − − ≈ × ~ ~ p power without any scattering. Each of Es and Bs∗ falls off ikr i~k ~x e i 2π ∞ 2 as 1/r, so the product falls off as 1/r2 and is negligible for ρdρdφ − · ρdρeikρ /2z 2 2 z large r.Thusif∆P is the change in the power of the Z z + ρ ≈ Z0 beam ( δP is the power lost), p 2π ∞ iku/z 2π − = du e = i . z Z k 1 ~ ~ ~ ~ 0 ∆P = ρdρdφRe Ei Bs∗ + Es Bi∗ . 2µ0 Z h × × iz

Thus Physics 504, Index of Refraction Physics 504, 2 Spring 2010 Spring 2010 π E0 ~ ~ ~ ~ ~ ~ ~ ~ Electricity Electricity ∆P = Re i~i f ∗(ki, ki)k + i~i kf ∗(ki, ki) and Consider thin slab, thickness d and ωµ 0k − · · Magnetism k Magnetism N i Shapiro Number density θ Shapiro ~ ~ ~ ~ ~ ~ ~ ~ i~k ~x x k R i~i ∗ f(ki, ki)ki i(ki f(ki, ki))~i ∗ E~ E ~ e i · − ·  incident wave i = 0 i · , ρ z Optical Optical ~ki = keˆz, observe from far z 2 Theorem and 0 d z0 Theorem and π E0 ~ ~ ~ ~ ~ ~ Index of z Index of = Re i~i f ∗(k, ki)+0+i~i ∗ f(ki, ki) 0 Refraction downstream, 0. Refraction 3 3 ωµ0 − · · − wavefront   The Optical Each d x in slab has Nd x scat- The Optical 2 Theorem Theorem 2π E0 ~ ~ ~ terers, so = Im ~i ∗ f(ki, ki) . Index of Index of − ωµ 0 · Refraction Refraction   ikR e i~k ~x 3 dE~ = f~(k, θ, φ;~k )E e i· Nd x The power flux in the incident beam is s R i 0 d 2π E 2 E 2k ~ ikz ∞ 1 ~ ~ 0 ~ 0 Es = NE0 dze dφ ρdρ Re (Ei Bi∗)z = | | Re ~i k ~i = | | Z Z Z 2µ0 × 2ωµ0 × × z 2ωµ0 0 0 0    ikR e 1 z0 z so the total cross section must be f~ k, cos− − ,φ; keˆ R   R  z 2∆Pωµ0 4π ~ ~ ~ σTot = − 2 = Im ~i ∗ f(ki, ki) . 2 2 2 E0 k k · As R = ρ +(z z) , ρdρ= RdR,so | |   0 − This is the optical theorem. Physics 504, Dropping that term, we have Physics 504, Spring 2010 Spring 2010 Electricity Electricity ikR d 2π ∞ e z0 z and NE0 and ~ 1 Magnetism ikz ik(z0 z) Magnetism ρdρ f k, cos− − ,φ; keˆz E~s = i dze dφe − f~(k, 0,φ; keˆz) Z0 R   R   k Z0 Z0 Shapiro Shapiro NE d ∞ ikR 1 z0 z 0 ikz0 ~ = dR e f~ k, cos− − ,φ; keˆz =2πi e f (k, 0, 0; keˆz) R Optical k Optical Z z0 z     Theorem and Theorem and | − | Index of Index of 1 ikR 1 z0 z ∞ Refraction Thus the total electric field at points far beyond the slab Refraction = e f~ k, cos− − ,φ; keˆz ik   R   The Optical is The Optical R= z0 z Theorem Theorem | − | ~ ikz 2πiNd ~ 1 ∞ ikR d 1 z0 z Index of E(~x)=E0e ~i + f(k, 0) , Index of e dR f~ k, cos− − ,φ; keˆz Refraction  k  Refraction −ik Z z0 z dR   R   | − | This is a plane wave, and exact solution of the free space where we integrated by parts for the last expression. The wave equation, though with shifted phase, amplitude, and last term is polarization. Thus it holds right up to back edge of the slab. 1 ∞ ikR z0 z d e dR − f~(k, θ, φ; keˆz) What was the effect of the slab? Project on original ik Z R2 d cos θ z0 z polarization — initial ~ ∗ E~ has been multiplied by | − | i · which, provided the indicated derivative is not singular, 1 1+2πik− N~i ∗ f~(k, 0)dz. falls off like 1/R. ·

Physics 504, Physics 504, Spring 2010 Caveats Spring 2010 Electricity Electricity and and Magnetism We assumed each scatterer feels only the incident field. Magnetism ~ ~ Integrating for finite thickness, Shapiro Better treatment says evaluate f(k) at the wavenumber in Shapiro the medium, not vacuum. 1 ~ ~ 2πik− N~i ∗ f(k,0)z ikz Optical Optical ~i ∗ E(~x)=e · E0e , Theorem and Theorem and · Index of Absorption will give imaginary part to Index of Refraction Refraction k nk k nk k i αk That is, our wave has replaced by , with The Optical = i =Re + 2 i with The Optical Theorem Theorem ~ 2πN~i ∗ f(k, 0) Index of 4πN ~ ~ ~ Index of n =1+ · , Refraction α = Nσtot = Im ~i ∗ f(k, k) . Refraction k2 k  ·  which is the index of refraction. Only full f~ will satisfy the optical theorem. Conclusion: The index of refraction is given by the Approximations suitable for Im f in the forward direction forward scattering amplitude. may not work for σ f 2 . tot ∝| | Example: small lossless dielectric sphere, f is real! So there is scattering but zero total cross section — can’t be. In 7.10D Kramers-Kronig said Re  1 is given by dω § r − 0 of Im r(ω0), so a purely real r can only be 1. R